Colimit of submodules

Question

I was going through a proof in the paper "Local unit versus local projectivity", where I came across the fact that for an $R$ module $P$ if $P = \operatorname{colim}\limits_{i\in I} P_i$ where $I$ is a directed set and $P_i$'s are finitely generated projective submodules of $P$ then for each $x \in P$ there exists some $k \in I$ such that $x \in P_k$. I don't have a very good knowledge on colimit in category theory. Any help in this matter would be highly appreciated.

Answer 1

Let $R$ be a commutative ring, $I$ be a directed set and $\varphi_{ij}:M_i\to M_j$ with $i,j\in I$ and $i\leq j$ be a diagram of $R$-modules that's $\varphi_{ii}=\mathrm{id}|M_i$ and $\varphi_{jk}\circ\varphi_{ij}=\varphi_{ik}$ for $i\leq j\leq k$. Let $\displaystyle M\cong\lim_\rightarrow M_i$ be a colimit with co-cone $\varkappa_i:M_i\to M$ so that $\varkappa_i=\varkappa_j\circ\varphi_{ij}$ for $i\leq j$. Then $M$ is union of the images of $\varkappa_i$, that's $$M=\bigcup_{i\in I}\mathrm{Im}(\varkappa_i)$$

For prove this, let $N=\bigcup_{i\in I}\mathrm{Im}(\varkappa_i)$. Then $N$ is a $R$-submodule of $M$, for if $x,y\in N$ then $x=\varkappa_i(x')$ and $y=\varkappa_j(y')$ for some $x'\in M_i$ and $y'\in M_j$ with $i,j\in I$. Since $I$ is directed, there exists $k\in I$ such that $i\leq k$ and $j\leq k$. Consequently, $x=\varkappa_i(x')=\varkappa_k\circ\varphi_{ik}(x')$ and $y=\varkappa_j(y')=\varkappa_k\circ\varphi_{jk}(y')$ hence $x,y\in\mathrm{Im}(\varkappa_k)$, thus are defined $x\pm y\in N$ and $ax\in N$ for $a\in R$ by means of $R$-module structure on $\mathrm{Im}(\varkappa_k)\subseteq N$.

In order to prove $M=N$ consider the projection $\pi:M\to M/N$ and for $a\in R$ the homothety $\omega_a:x\in M/N\mapsto ax:M/N\to M/N$. Then $\omega_a\circ\pi\circ\varkappa_i=0=\pi\circ\varkappa_i$ for each $i\in I$ hence by universal property (uniqueness) $\omega_a\circ\pi=\pi$, that's $ax=x$ for each $a\in R$ and $x\in M/N$, that's $(a-1)x=0$. Thus $x=(a-1)x=0$ for each $x\in M/N$ and this proves $M/N=\{0\}$ that's $M=N$ as required.